Question

if cos$\theta$+sin$\theta$=$\sqrt{2}$ cos$\theta$, show that cos$\theta$ - sin$\theta$= $\sqrt{2}sin[tex]\theta$ ​

Answer 1

--> cos θ + sin θ = √2 cos θ

--> sin θ = √2 cos θ - cos θ

=> sin θ = ( √2 - 1 ) cos θ

=> [ sin θ / ( √2 - 1 ) ] = cos θ

=> [ sin θ ( √2 + 1 ) / ( 2 - 1 ) ] = cos θ

=> [ √2 sin θ + sin θ ] = cos θ

=> cos θ - sin θ = √2 sin θ

Answer 2

$\boxed{\bold{\huge{SOLUTION\:\: :}} }$

HERE, I AM HOLDING "THETAS" AS "X".

Then,

cos X + sin X = √2 cos X.

We can definitely say,

cos²X + sin²X = 1

Or, (cosX + sinX)² - 2 cosX sinX = 1

Or, (√2 cosX)² = 1 + 2 cosX sinX

Or, 2 cos²X = sin²X + cos²X + 2 cosX sinX

Or, cosX sinX = (cos²X - sin²X)/2

Now, (cosX - sinX)²

= 1 - 2 sinX cosX

= 1 - 2 {(cos²X-sin²X)/2}

= 1 - cos²X + sin²X

= 2 sin²X

HENCE, WE CAN SAY :

(cosX - sinX) = √2 sin X [PROVED]